[SOLVED] Probably String Parsing or Something

jroggow · 01-25-2013, 08:08 PM

I'm trying to improve my bash scripting skills. Presently my scripts look like a brick shithouse. They work, but they ain't pretty.

My practice project is converting all of my music files to .ogg format and moving them into a dedicated directory.

I have a script to convert the files to .ogg and move them to my music directory, but I'm stumbling over a few details.

My script creates files named something like getbent.wav.ogg. Which is a bit ugly, so the move part of my script changes the file name to 'NUMBER'.ogg. NUMBER increments with each iteration so my music directory has 0.ogg, 1.ogg, 2.ogg, &c.

I'm fine with such a naming scheme since I don't need to care about the actual file name. I only need to care that my files aren't being overwritten. Given that my file's names are all numbers, I should be able to find the highest increment assigned to a file in the directory, assign my variable that value, plus one, and then continue the script as normal.

I can't for the life of me figure out how to find the highest increment in the directory and start the script anew from there. I could count the items in the directory listing, but since there is a significant possibility I will accidentally save something unrelated there, I would rather find the highest number in the file name of .ogg format.

How do I strip the extension and convert the remainder to an integer? And do I need to do that or am I over-thinking myself? Why are my socks wet? Have you seen my underpants? When will these questions end?

Anybody? Buhler?

jlinkels · 01-25-2013, 08:24 PM

You want to find the highest number first:

Code:

lastfile=$(ls *.ogg | sort -g | tail -n 1)

Then strip the extension:

Code:

lastnum=${lastfile%.ogg}

You don't need to convert to integer, Bash is typeless.

You do need the sort -g. In ls 10.ogg comes before 3.ogg.

jlinkels

jroggow · 01-25-2013, 08:44 PM

Curly braces are my new best friend. I was tinkering with regular expressions and SED.

danielbmartin · 01-25-2013, 09:07 PM

Radioactive chickens will teach you how to spell Bueller.

See http://www.youtube.com/watch?v=uhiCFdWeQfA

Daniel B. Martin

David the H. · 01-27-2013, 01:53 PM

Uggh, not ls again.

Code:

files=( *.ogg )
highest=${files[-1]}	#use ${files[@]:(-1)} in older bash versions

number="${highest%.*}"	#to simply remove the file extension

re='([0-9]+)'           #regex extraction of a number from a longer filename
[[ $highest =~ $re ]]
number=${BASH_REMATCH[1]}

However, this does require that the numbers are all zero-padded so that they automatically sort properly in the shell. Otherwise it gets more complex. I strongly recommend always using zero-padded numbers in filenames for this reason.

You can find a script I wrote that zero-pads file numbers in this thread:
http://www.linuxquestions.org/questi...rectory-949189

However, also be warned that the shell's arithmetic engine treats numbers with leading zeros as octal values, which will give you errors and bizarre-seeming outputs if you aren't aware of it. Check out this this page for more on how to handle that:

arithmetic expressions

See here too for more on using zero-padded numbers:
http://mywiki.wooledge.org/BashFAQ/018

And much more on bash's built-in string manipulation features here:
http://mywiki.wooledge.org/BashFAQ/100

Edit: another way to get the largest number, if not zero-padded:

Code:

greatest=0
for fname in *.ogg; do
   (( ${fname%.*} > ${greatest%.*} )) && greatest=$fname
done

echo "${greatest%.*}"


re='([0-9]+)' gtnum=0

for fname in *.ogg; do
    [[ $fname =~ $re ]] && fnum=${BASH_REMATCH[1]}
    (( fnum > gtnum )) && greatest=$fname && gtnum=$num
done

echo "$greatest ; $gtnum"